Problem 18 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R}\left(\frac{\sqrt{x}}{y^{2}}\right) d A, \quad R : \quad 0 \leq x \leq 4, \quad 1 \leq y \leq 2$$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{8}{3}\).

Step 1: Set Up the Double Integral

The integral given is \(\iint_{R}\left(\frac{\sqrt{x}}{y^{2}}\right) dA\), where the region \(R\) is defined as \(0 \leq x \leq 4\) and \(1 \leq y \leq 2\). We need to integrate first with respect to \(x\) and then with respect to \(y\). The expression can be rewritten as: \[ \int_{1}^{2} \int_{0}^{4} \frac{\sqrt{x}}{y^{2}} \, dx \, dy. \]

Step 2: Integrate with Respect to x

First, integrate \(\frac{\sqrt{x}}{y^{2}}\) with respect to \(x\) from 0 to 4:\[\int_{0}^{4} \frac{\sqrt{x}}{y^{2}} \, dx = \frac{1}{y^{2}} \int_{0}^{4} \sqrt{x} \, dx.\]Calculate \(\int_{0}^{4} \sqrt{x} \, dx\): \[\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3} x^{3/2} + C.\]Evaluate from 0 to 4: \[\frac{2}{3} [4^{3/2} - 0^{3/2}] = \frac{2}{3} \times 8 = \frac{16}{3}.\]

Step 3: Integrate with Respect to y

Now integrate the result \(\frac{16}{3y^{2}}\) with respect to \(y\):\[\int_{1}^{2} \frac{16}{3y^{2}} \, dy = \frac{16}{3} \int_{1}^{2} y^{-2} \, dy.\]Calculate: \[\int y^{-2} \, dy = -y^{-1} = -\frac{1}{y} + C.\]Evaluate from 1 to 2: \[\left[-\frac{1}{y}\right]_{1}^{2} = \left(-\frac{1}{2}\right) - \left(-1\right) = -\frac{1}{2} + 1 = \frac{1}{2}.\]Thus, multiply by \(\frac{16}{3}\): \[\frac{16}{3} \times \frac{1}{2} = \frac{16}{6} = \frac{8}{3}.\]

Step 4: Conclusion

The evaluation of the double integral over the given region \(R\) is \(\frac{8}{3}\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrating functions. It allows us to find quantities like areas, volumes, and accumulations.
In the exercise above, we look at a double integral, which is a specific type of integral that computes a volume under a surface in a two-dimensional region.
The function \( \frac{\sqrt{x}}{y^{2}} \) is integrated over a set region defined with boundaries. This operation helps in determining a cumulative quantity over the specified area.

• In single-variable calculus, we deal with one-dimensional intervals; however, in this double integral, our interval extends into two dimensions (over the area \( R \)).
• We utilize double integrals to calculate the area, much like two single integrals but conducted over two different intervals.

Understanding how to set up and compute double integrals is important in solving practical problems involving areas and volumes.

Area of a Region

When evaluating double integrals, the area of the region \( R \) is crucial. This is the domain over which the function is integrated.
For this exercise, the region \( R \) is given by \{(x,y) | 0 \leq x \leq 4, \, 1 \leq y \leq 2\}. This region is a rectangle in the \(xy\)-plane.
Understanding the region helps in setting correct limits for integration.

• The rectangle has a width along the \(x\)-axis from 0 to 4 and a height along the \(y\)-axis from 1 to 2.
• This geometry aids in determining how to iterate the integration, which variable to integrate first, impacting setup and solution.

The specified area \( R \) is a common starting point for computing double integrals, providing the necessary scope for integration.

Integration Limits

Integration limits define the start and end points for integration over a specified region. In double integrals, each variable has its own limits. Here, \( x \) goes from 0 to 4 and \( y \) from 1 to 2.
These limits frame the problem and guide the integration process.

• They determine the boundaries for each part of the area \( R \).
• By setting limits properly, we ensure the integration evaluates the function over the correct domain.
• The iterated integral structure involves two stages: integrating with respect to \( x \) within its limits first, followed by \( y \).

Correct setup of integration limits is essential to ensuring the accuracy of the computed double integral.

Iterated Integrals

Iterated integrals involve performing integration in stages, first one variable then the other. Here, the process starts with integrating \( \frac{\sqrt{x}}{y^{2}} \) with respect to \( x \) before \( y \).
This sequential process highlights the heart of the double integral calculation.

• The function is evaluated by removing one variable, simplifying the integration and breaking it down into manageable parts.
• The inner integral \[ \int_{0}^{4} \frac{\sqrt{x}}{y^{2}} \, dx \] is evaluated first, producing an expression in terms of \( y \).
• The outer integral \[ \int_{1}^{2} \frac{16}{3y^{2}} \, dy \] completes the process by integrating this result over \( y \), giving a full solution.

By iterating integrals, solving these stepwise allows for a detailed approach that contributes to understanding more complex integrative processes.